Residue of $f(z) = \frac{e^{3z}}{(z+1)^2}$

Question

I want to calculate the residue of $f(z) = \frac{e^{3z}}{(z+1)^2}$ around $z_0 = -1$

I can see that there is a pole of degree 2 at -1. My thought is to use the limit for Res but I'm not pretty sure because of the numerator.

Here is what I've done :

$$\text{Res}_{z=-1} f(z) = \lim_{z\to -1} \frac{d}{dz}[(z+1)^2f(z)] = \lim_{z\to-1} \frac{d}{dz}e^{3z} = 3e^{-3}$$

Is my thought correct? Or I can't calculate it this way?

Answer 1

Your thoughts are correct. The residue $= 3e^{-3}.$